Find the area of the region bounded by the graphs of the equations f(x)=-x^(2)+4x and y=0

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We are asked to find the area of the region bounded by `f(x)=-x^2+4x` and `y=0` .

We can factor the quadratic to y=-x(x-4); the x-intercepts are 0 and 4.

The equivalent question then is to find the definite integral `int_0^4 (-x^2+4x)dx` .

`int_0^4 (-x^2+4x)dx=-x^3/3+2x^2 |_0^4`

`=(-64/3+32)-(0+0)=32/3 `

So the area...

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We are asked to find the area of the region bounded by `f(x)=-x^2+4x` and `y=0` .

We can factor the quadratic to y=-x(x-4); the x-intercepts are 0 and 4.

The equivalent question then is to find the definite integral `int_0^4 (-x^2+4x)dx` .

`int_0^4 (-x^2+4x)dx=-x^3/3+2x^2 |_0^4`

`=(-64/3+32)-(0+0)=32/3 `

So the area bounded by the given curve and the x-axis is 32/3.

An alternative approach is to use a Riemann sum:

The area can be found by `lim_{n->oo} sum_{i=1}^{n} f(c_{i})Delta x_{i}`

If we use a regular partition and choose the right-hand endpoint in each subinterval we get:

`A=lim_{n->oo} sum_{i=1}^{n} (-((4i)/n)^2+4((4i)/n))(4/n) `

`=lim_{n->oo}[(-64)/(n^2)sum_{i=1}^n(i^2/n-i)] `

`=lim_{n->oo}[(-64)/n^2(1/n((n(n+1)(2n+1))/6)-(n(n+1))/2)] `

`=lim_{n->oo}[(-64)/n^2(1/3n^2+1/2n+1/3-n^2/2-n/2)] `

`=lim_{n->oo}[(-64)/3-(32)/n-(64)/n^3+32+(32)/n] `

`=(-64)/3+32=32/3=10.bar(6) `

The graph:

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